Answer
$$e^{-1}$$
Work Step by Step
$(\dfrac{1}{n})^{1/\ln n}=(e^{ln (1/n))^{1/\ln n}}\\=(e^{-\ln n })^{1/\ln n }\\=e^{-1}$
Calculus: Early Transcendentals (2nd Edition)
by Briggs, Bill L.; Cochran, Lyle; Gillett, Bernard
Published by
Pearson
ISBN 10:
0321947347
ISBN 13:
978-0-32194-734-5
Chapter 8 - Sequences and Infinite Series - Review Exercises - Page 658: 7
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